Problem: What is the modular inverse of $11$, modulo $1000$?

Express your answer as an integer from $0$ to $999$, inclusive.
Solution: We know that the modular inverse exists because $11$ and $1000$ are relatively prime. Notice that $1000 = 10^3$ and that $11 = 10 + 1$. Since $11 \cdot 11^{-1} \equiv 1 \pmod{1000}$, it follows that $(10+1) \cdot 11^{-1} = 10^3k + 1$ for some integer $k$. We recognize the potential sum of cubes factorization: if $k=1$, then $$10^3 + 1 = 1001 = 11 \cdot (10^2 - 10 + 1) = 11 \cdot 91.$$Thus, $11^{-1} \equiv \boxed{91} \pmod{1000}$.